Determine the pressure at a depth of 0.6 m in water, kerosene, mercury, calculate the water pressure

| July 13, 2021 | education

Determine the pressure at a depth of 0.6 m in water, kerosene, mercury, calculate the water pressure at the bottom of one of the deepest sea trenches, which is at a depth of 10900 m, the density of seawater is 1030 kg / m 3

1) The pressure of the liquid column is calculated by the formula:

P = pgh, where P is the pressure, p is the density of the liquid, g is the coefficient of free fall (in SI = 9.8 m / s), h is the height of the liquid column.

2) According to the table of densities in the textbook, we find out the densities of liquids from the condition of the problem:

p (water) = 1000 kg / m³.

p (kerosene) = 800 kg / m³.

p (mercury) = 13600 kg / m³.

p (sea water) = 1030 kg / m³.

3) Find the pressure:

P (in water) = 1000 kg / m³ * 9.8 m / s * 0.6 m = 5800 Pa.

P (in kerosene) = 800 kg / m³ * 9.8 m / s * 0.6 m = 4704 Pa.

P (in mercury) = 13600 kg / m³ * 9.8 m / s * 0.6 m = 79968 Pa.

P (at the bottom of the Mariana Trench) = 1030 kg / m³ * 9.8 m / s * 10900 m = 110024600 Pa.

Answer: 5800 Pa, 4704 Pa, 79968 Pa, 110024600 Pa.



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